Question

1. The octane rating of gasoline is a relationship of the burning efficiency of the given gasoline mixture to the burning efficiency of octane (C₈H₁₈). Like most of hydrocarbons, octane reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reaction:

                        C₈H₁₈ (l) + O₂(g) → CO₂ (g) + H₂O (g)

If 0.240 mol of octane is allowed to react with 0.890 mol of oxygen gas,

1. how many moles of water are produced in this reaction (theoretical yield of water)?
2. How much excess reactant (E.R.) is left in moles?

Answer 1

The combustion of octane occcurs as follows:
2C₈H₁₈ (l) + 25O₂(g) → 16CO₂ (g) + 18H₂O (g)
It is evident from the balanced equation that:
2 mols of C₈H₁₈ require 25 mol of O₂ to react completely.
1 mol of C₈H₁₈ requires 12.5 mol of O₂ to react completely.
0.240 mol of C₈H₁₈ requires O₂ = 12.5 x 0.240 = 3
However, only 0.89 mol of O₂ is available.
Hence, O₂ is the limiting reagent and C₈H₁₈ is the excess reagent.
The quantity of product formed depends on the quantity of the limiting reagent.
25 mol of O₂ to react to produce H₂O = 18 mol
1 mol of O₂ to react to produce H₂O = 18/25 mol
0.89 mol of O₂ to react to produce H₂O = (18/25) x 0.89 = 0.6408 mol
0.89 mol of O₂ reacts with C₈H₁₈ = (2/25) x 0.89 = 0.0712 mol
Number of moles of (excess reagent) left unreacted = 0.240 -0.0712 = 0.1688 mol